求积分：∫dx/sin2x+2sinx

题目

求积分：∫dx/sin2x+2sinx
需过程,谢谢!

答案

1/[sin2x+2sinx]
=1/[2sinxcosx+2sinx]
=1/[2sinx(1+cosx)]（上下都乘以sinx）
=sinx/[2sinx*sinx*(1+cosx)]
所以
∫dx/sin2x+2sinx
=1/2∫sinx/[(1-(cosx)^2)(1+cosx)]dx
=-1/2∫1/[(1-(cosx)^2)(1+cosx)]dcosx(凑微分法,记cosx=t)
=-1/2∫1/[(1-t^2)(1+t)]dt
=-1/2{-1/4*ln(t-1)-1/2*1/(1+t)+1/4*ln(1+t)}+C
=1/8*(ln(cosx-1)+ln(cosx-1)*cosx+2-ln(1+cosx)-ln(1+cosx)*cosx)/(1+cosx)+C